Define function with nonlinear equation system vercat error
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Hi,
I am trying to define a nonlinear equation system in a function in order to solve it using fsolve.
Already calling the function it self raises the error
"Error using vertcat
Dimensions of arrays being concatenated are not consistent."
running fminunc results in
Error in fminunc (line 307)
f = feval(funfcn{3},x,varargin{:});
Error in GPS_Calculation (line 49)
sol = fminunc(f,[6 6 6 6])
Caused by:
Failure in initial objective function evaluation. FMINUNC cannot continue.
How can I fix this?
f = @(x)[sqrt( (101 - x(1)).^2 + (16 - x(2)).^2 + (207 - x(3)).^2 ) + x(4) - 310.5685;
sqrt( (52 - x(1)).^2 + (21 - x(2)).^2 + (302 - x(3)).^2 ) + x(4) - 387.5097;
sqrt( (17 - x(1)).^2 + (53 - x(2)).^2 + (350 - x(3)).^2 ) + x(4) -434.7066;
sqrt( (-15 - x(1)).^2 + (159 - x(2)).^2 + (208 - x(3)).^2 ) + x(4) - 341.25730]
f([6 6 6 6])
sol = fminsearch(f,[6 6 6 6])
sol = fminunc(f,[6 6 6 6])
sol = fsolve(F,[6 6 6 6]
Answers (2)
F = @(x)[sqrt( (101 - x(1)).^2 + (16 - x(2)).^2 + (207 - x(3)).^2 )+ x(4)- 310.5685;
sqrt( (52 - x(1)).^2 + (21 - x(2)).^2 + (302 - x(3)).^2 )+x(4)-387.5097;
sqrt( (17 - x(1)).^2 + (53 - x(2)).^2 + (350 - x(3)).^2 )+x(4)-434.7066;
sqrt( (-15 - x(1)).^2 + (159 - x(2)).^2 + (208 - x(3)).^2 )+x(4)-341.25730];
f=@(x) norm(F(x))^2;
[sol,fval] = fminsearch(f,[6 6 6 6],optimset('TolFun',1e-12','MaxIter',1e5,'MaxFunEvals',inf))
opts=optimoptions('fminunc','StepTol',1e-12,'OptimalityTol',1e-12,'FunctionTol',1e-12,'Display','none');
[sol, fval] = fminunc(f,[6 6 6 6],opts)
opts=optimoptions('fsolve','StepTol',1e-12,'OptimalityTol',1e-12,'FunctionTol',1e-12,'Display','none');
[sol,fval] = fsolve(F,[6 6 6 6],opts)
7 Comments
weggee
on 24 Jan 2022
Thank you!
You're welcome, but please Accept-click the answer if your question is settled.
Can you elaborate why calling norm() is necessary?
fminsearch and fminunc are function minimizers. It is not possible to minimize a vector-valued quantity. You can only do a comparison a>b when a and b are scalars.
Torsten
on 24 Jan 2022
Alternatively, you can use "fsolve" to solve 4 equations in 4 unknowns:
f = @(x)[sqrt( (101 - x(1)).^2 + (16 - x(2)).^2 + (207 - x(3)).^2 ) + x(4) - 310.5685;
sqrt( (52 - x(1)).^2 + (21 - x(2)).^2 + (302 - x(3)).^2 ) + x(4) - 387.5097;
sqrt( (17 - x(1)).^2 + (53 - x(2)).^2 + (350 - x(3)).^2 ) + x(4) - 434.7066;
sqrt( (-15 - x(1)).^2 + (159 - x(2)).^2 + (208 - x(3)).^2 ) + x(4) - 341.25730]
xsol = fsolve(f,[6 6 6 6])
weggee
on 24 Jan 2022
Walter Roberson
on 24 Jan 2022
You have four independent goals, and it might not be possible to satisfy all four of them simultaneously.
There is no way to put them all together into one equation without deciding how you want to weight the relative importance of each of them being satisfied.
Matt J
on 24 Jan 2022
You caould have done
f = @(x)norm( [sqrt( (101 - x(1)).^2 + (16 - x(2)).^2 + (207 - x(3)).^2 )+ x(4)- 310.5685;
sqrt( (52 - x(1)).^2 + (21 - x(2)).^2 + (302 - x(3)).^2 )+x(4)-387.5097;
sqrt( (17 - x(1)).^2 + (53 - x(2)).^2 + (350 - x(3)).^2 )+x(4)-434.7066;
sqrt( (-15 - x(1)).^2 + (159 - x(2)).^2 + (208 - x(3)).^2 )+x(4)-341.25730] );
Walter Roberson
on 24 Jan 2022
0 votes
You have a multi objective search, trying to simultaneously minimize four different objectives. fmincon is only able to minimize a single objective. You need to switch to a Pareto search using gamultiobj() or paretosearch()
Remember that Pareto searches are not global minima searches: they correspond to finding local minima such that moving the point in any direction makes at least one of the objectives worse.
1 Comment
There happens to be a unique solution. But notice that I did not use fmincon()
syms x [1 4]
eqn = [sqrt( (101 - x(1)).^2 + (16 - x(2)).^2 + (207 - x(3)).^2 ) + x(4) - 310.5685;
sqrt( (52 - x(1)).^2 + (21 - x(2)).^2 + (302 - x(3)).^2 ) + x(4) - 387.5097;
sqrt( (17 - x(1)).^2 + (53 - x(2)).^2 + (350 - x(3)).^2 ) + x(4) - 434.7066;
sqrt( (-15 - x(1)).^2 + (159 - x(2)).^2 + (208 - x(3)).^2 ) + x(4) - 341.25730]
sol = solve(eqn)
format long g
X1 = double(sol.x1)
X2 = double(sol.x2)
X3 = double(sol.x3)
X4 = double(sol.x4)
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